(i) Note that t = 2π/6 <==> the point (cos 2π/6, sin 2π/6, 2π/6) = (1/2, sqrt(3)/2, π/3).
(ii) For the direction vector:
Differentiating the curve yields x' = -sin t, y' = cos t, z' = 1.
So, the direction vector v for the tangent line is (x', y', z') at t = 2π/6 (where the point is located):
==> v = (-sin 2π/6, cos 2π/6, 1) = (-sqrt(3)/2, 1/2, 1).
Hence, the equation of the tangent line at the prescribed point is
r(t) = (1/2, sqrt(3)/2, π/3) + t(-sqrt(3)/2, 1/2, 1).
Parametrically, this is given by
x = 1/2 - t sqrt(3)/2
y = sqrt(3)/2 + t/2
z = π/3 + t.
I hope this helps!